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Some curves defined by polynomial equations are disconnected over reals but not over complexes, e.g., $x y - 1 = 0$. How can we convince someone with background only on equations over reals that the curve drawn by above equation is connected over complexes? Is a plot or something possible, for example? It will be a 4d plot if x and y are expanded to real and imaginary parts. Any other plot, or algebraic way to show connectedness?

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    $\begingroup$ @AlexL Even if your post wasn't exactly what the OP was looking for, I think it was a very good way of seeing the solution. Please consider un-deleting your answer. $\endgroup$ – André 3000 Oct 22 '18 at 4:18
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A solution over $\mathbb{C}$ is a pair $(z, 1/z)$ with $z \neq 0$. Consider another solution $(w, 1/w)$. There is a path from $z$ to $w$ in $\mathbb{C}$ which does not cross $0$ (this is the difference with $\mathbb{R}$), and this yields a path from $1/z$ to $1/w$ by inverting every point on the path. From this we get a path from $(z, 1/z)$ to $(w, 1/w)$ which lives inside the solution set.

You can draw the corresponding plots explicitly. It is especially easy on the unit circle, where inversion is just complex conjugation. For example, to go from $-1$ to $1$ in $\mathbb{C}$ you can take the upper semicircle of the unit circle. The "inverted" path (inverting every complex number on the path) is exactly the lower semicircle. This shows how to go from $(-1,-1)$ to $(1,1)$ (which wasn't possible over the reals) via the complex domain.

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  • $\begingroup$ Good answer, but it is still a bit hand-waiving in the end than what I was looking for. How would one plot this curve? It is still poits (x,1/x) in complex dimension, so we have to plot it in 4D, right? Not sure I completely understand the last paragraph. $\endgroup$ – dbm Oct 22 '18 at 13:14
  • $\begingroup$ I identify a path in $\mathbb{C}^2$ with a pair of paths in $\mathbb{C}$ (in the same way one can identify a path in $\mathbb{R}^2$ with two paths in $\mathbb{R}$); one can view it as projection to the axes (complex axes here). The last paragraph has an explicit example of such a pair of paths. These two paths (which you can draw in the same copy of $\mathbb{C}$) are two different views/shadows of a path in $\mathbb{C}^2$ (views along two different complex axes), which together determine the path in $\mathbb{C}^2$ uniquely. $\endgroup$ – Ricardo Buring Oct 22 '18 at 14:49
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You can connect any two complex points $u, v$ with a spiral

$$u^{1-t}v^t$$ where $t$ runs from $0$ to $1$.

More precisely, in parametric polar coordinates,

$$\begin{cases}\theta=\theta_u(1-t)+\theta_vt,\\\rho=\rho_u^{1-t}\rho_v^t.\end{cases}$$

(You can very well eliminate $t$.)

Then the curve $(-\theta,\rho^{-1})$ corresponds to the inverse points and is a similar continuous spiral.

enter image description here

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  • $\begingroup$ what is the direct relation with the example system I have in the question? $\endgroup$ – dbm Oct 22 '18 at 13:16

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